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I was preparing a class on polynomials (high school level). The handbook I use always contains some questions from math olympiads. The following question is asked:

What is the number of positive integer solutions of the following inequality: $$(x-\frac{1}{2})^1(x-\frac{3}{2})^3\ldots(x-\frac{4021}{2})^{4021} < 0.$$

I found this number to be 1005. Here is my reasoning:

For a positive integer to satisfy this inequality, an odd number of factors should be negative. There are $\frac{4021 +1}{2} =2011$ distinct factors. Each distinct factor appears an odd number of times, hence we can forget about the exponents. We also only need to consider positive integers smaller than $\frac{4021}{2} = 2010,5$, for any bigger number would make all factors positive.

In order for the product to be negative, we need to look for positive integers making an even amount of the first factors positive. This happens if we pick even integers: the constants $\frac{1}{2}, \frac{3}{2},...$ are one unit from the next, so if we pick an even integer, there are an even amount of such constants smaller than that integer.

Hence we look for the number of positive integers smaller than $2010,5$, which is $\frac{2011 - 1}{2} = 1005$.

Is this solution correct? (sorry for the bad english by the way. I hope this does not make my explanation unclear).

Note I am also interested in other (quicker, slicker) solutions.

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This may look shorter, but then again it is not really different from your argument:

The polynomial changes its sign at $\frac12, \frac32,\ldots, \frac{4021}2$ and is positive as $x\to+\infty$, hence positive for $x>\frac{4021}2$. In the $2010$ intervals determined this way, the polynomial is alternatingly positive and negative, hence negative in $1005$ of them. Each such interval contains exactly one positive integer (whereas all other positive integers are $>\frac{4021}2$), hence the answer is $1005$.

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  • $\begingroup$ Thank you! .... $\endgroup$ – Student Oct 18 '18 at 20:12

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